ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A biased forward solve shifts a Gaussian posterior

Example

Let the scalar parameter $u$ have prior density $p_{0} (u) = (2 π)^{- 1 / 2} e^{- u^{2} / 2},$ and observe $y = 0$ with unit-variance Gaussian noise. The exact forward map is $G (u) = u$ , while a deliberately biased numerical forward/PDE solver reports $\tilde{G} (u) = u + 2$ . At $y = 0$ , use likelihoods $L (u) = \exp (- G (u)^{2} / 2)$ and $\tilde{L} (u) = \exp (- \tilde{G} (u)^{2} / 2)$ ; the common Gaussian density constant cancels when normalizing. Define each fixed-data posterior by normalizing the prior times its stated likelihood. The exact posterior is $N (0, 1 / 2)$ and the solver posterior is $N (- 1, 1 / 2)$ . For the bounded functional $f (u) = 1_{(- \infty, 0]} (u)$ , their expectations are respectively $Π (f) = \frac{1}{2}, \tilde{Π} (f) = Φ_{0} (\sqrt{2}) > \frac{1}{2},$ where $Φ_{0} (t) : = \int_{- \infty}^{t} (2 π)^{- 1 / 2} e^{- z^{2} / 2} d z .$ The discrepancy is in the forward-solver stage of the posterior error budget; this toy solver assumption makes no claim about the bias of any general PDE discretization.

Example references

The model, forward-solver, and discretization stages of the error budget.

The standard normal density and affine normal laws.

Bounded expectations differ by at most $2 ∥ f ∥_{\infty}$ times library total variation.

Facts & Assumptions

Given: The prior density, the exact and solver forward maps, the unit-variance observation noise, and the normalized prior-times-likelihood posterior convention in the Example.

[F1]

The error-budget solver term compares the exact-model posterior expectation with the posterior expectation obtained from the declared numerical forward solve. Posterior computation error budget

[F2]

The standard normal density is $(2 π)^{- 1 / 2} e^{- z^{2} / 2}$ , and $N (m, s^{2})$ is its affine image $m + s Z$ for $s > 0$ . Standard normal and normal laws

[A1]

AC is assumed by the library standard-normal law interface. The Axiom of Choice

[F3]

For a bounded measurable functional, the expectation difference is at most $2 ∥ f ∥_{\infty}$ times the event-supremum total variation distance. Bounded posterior functionals and stationary sample error

Verification

1.1A1F2givenalgebra

With the likelihoods stated in the Example, the exact prior-times-likelihood product is $(2 π)^{- 1 / 2} e^{- u^{2}}$ . The substitution $z = \sqrt{2} u$ and the normalized standard-normal integral in [F2] give $\int_{R} e^{- u^{2}} d u = \sqrt{π}$ , so this product has normalizing integral $(2 π)^{- 1 / 2} \sqrt{π} = 1 / \sqrt{2}$ . Its normalized density is therefore $π^{- 1 / 2} e^{- u^{2}}$ , which is $N (0, 1 / 2)$ by [F2]. For the solver output, the prior-times-likelihood product is $(2 π)^{- 1 / 2} e^{- u^{2} / 2 - (u + 2)^{2} / 2} = (2 π)^{- 1 / 2} e^{- (u + 1)^{2} - 1};$ its normalizing integral is $(2 π)^{- 1 / 2} e^{- 1} \sqrt{π} = e^{- 1} / \sqrt{2}$ , and its normalized density is $π^{- 1 / 2} e^{- (u + 1)^{2}} = N (- 1, 1 / 2)$ by [F2].

2.1F2step 1.1givenalgebra

The exact density is symmetric about zero, so its mass on $(- \infty, 0]$ is $1 / 2$ . Under the solver density, $Z = \sqrt{2} (U + 1)$ is standard normal, and $U \leq 0$ is equivalent to $Z \leq \sqrt{2}$ ; hence $\tilde{Π} (f) = Φ_{0} (\sqrt{2})$ . The standard normal density is positive and symmetric, so $Φ_{0} (\sqrt{2}) > Φ_{0} (0) = 1 / 2$ .

3.1F1F3step 2.1algebra∎

In the error ledger set $μ_{⋆} = μ_{m o d e l} = Π$ and $μ_{s o l v e r} = μ_{d i s c} = \tilde{Π}$ ; then the solver-stage term is $δ_{s o l v e r} = Φ_{0} (\sqrt{2}) - 1 / 2$ and the model and discretization terms vanish. Since $∥ f ∥_{\infty} = 1$ , [F3] also gives $δ_{s o l v e r} \leq 2 ∥ Π - \tilde{Π} ∥_{T V}$ .

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